= Solution
Use the standard complex <inner product> $\langle u,v\rangle=u^*v$, conjugate-linear in its first argument. A <unitary matrix> satisfies $U^*U=I$, and a <Hermitian matrix> satisfies $H^*=H$, where $*$ is conjugate transpose.
If $Hv=\lambda v$ with $v\ne0$, then
$$
\lambda\langle v,v\rangle=\langle v,Hv\rangle=\langle Hv,v\rangle=\overline\lambda\langle v,v\rangle.
$$
Positivity of $\langle v,v\rangle$ proves $\boxed{\lambda\in\mathbb R}$. If $Uv=\lambda v$, preservation of the <norm> gives $\|v\|^2=\|Uv\|^2=|\lambda|^2\|v\|^2$, so $\boxed{|\lambda|=1}$.
For <eigenvectors> $Hv=\lambda v$ and $Hw=\mu w$, the <Hermitian matrix> identity gives $\langle v,Hw\rangle=\langle Hv,w\rangle$. The already established reality of the <eigenvalues> therefore implies $(\mu-\lambda)\langle v,w\rangle=0$. Thus \b[distinct <eigenvalues> have orthogonal <eigenvectors>].
Back to article page
